296 Вычислить: 1) log2 24 1log2 72 2 log318-log3 72 3

$$\frac{log_2 24 - \frac{1}{2}log_2 72}{log_3 18 - \frac{1}{3}log_3 72} = \frac{log_2 24 - log_2 \sqrt{72}}{log_3 18 - log_3 \sqrt[3]{72}} = \frac{log_2(\frac{24}{\sqrt{72}})}{log_3(\frac{18}{\sqrt[3]{72}})} = \frac{log_2(\frac{24}{6\sqrt{2}})}{log_3(\frac{18}{2\sqrt[3]{9}})} = \frac{log_2(\frac{4}{\sqrt{2}})}{log_3(\frac{9}{\sqrt[3]{9}})} = \frac{log_2(\frac{2^2}{2^{\frac{1}{2}}})}{log_3(\frac{3^2}{3^{\frac{2}{3}}})} = \frac{log_2(2^{\frac{3}{2}})}{log_3(3^{\frac{4}{3}})} = \frac{\frac{3}{2}log_2 2}{\frac{4}{3}log_3 3} = \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{3}{2} \cdot \frac{3}{4} = \frac{9}{8}$$.

Ответ: 9/8